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Show that $\dbinom{2n}{n} + \dbinom{2n}{n-1} = \frac{1}{2} \dbinom{2n+2}{n+1}$

By induction, suppose that for some n its true, $\dbinom{2n}{n} + \dbinom{2n}{n-1} = \dbinom{2n+2}{n+1}$ by theorem, but, I don't know how to prove that $\dbinom{2n+2}{n+1} = \frac{1}{2} \dbinom{2n+4}{n+2}$

MJD
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Laura Merchán
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    If you write these binomials in terms of factorials and add through a common denominator, it will get you pretty far – imranfat Feb 16 '20 at 19:12
  • $\dbinom{2n}{n} + \dbinom{2n}{n-1} = \dbinom{2n+2}{n+1}$? Try this when $n=1$ – saulspatz Feb 16 '20 at 19:13
  • $\dbinom{2}{1} + \dbinom{2}{0} = 3 =\frac{1}{2} \dbinom{4}{2} =\frac{6}{2} =3$ – Laura Merchán Feb 16 '20 at 19:39
  • When I write these binomials in terms and use some properties $\dbinom{2n+2}{n} + \dbinom{2n}{n-1} =\frac{1}{2} \dbinom{2n+4}{n+2} \rightarrow \frac{(2n+2)!}{(n+1)!(n+1)!} = \frac{1}{2} \frac{(2n+4)!}{(n+2)!(n+2)!}$ so I obtain that $\frac{(2n+2)!}{(n+1)!(n+1)!} = \frac{2n+3}{(n+2)} \frac{(2n+2)!}{(n+1)!(n+1)!}$ I don't know for what common denominator I have to add – Laura Merchán Feb 16 '20 at 20:13

2 Answers2

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$$\begin{align} {2n\choose n}+{2n\choose n-1}&=\frac{(2n)!}{n!n!}+\frac{(2n)!}{(n-1)!(n+1)!} \\ &= \frac{(2n)!(n+1)}{n!(n+1)!}+\frac{(2n)!n}{n!(n+1)!} \\ &= \frac{(2n)!(n+1+n)}{n!(n+1)!} \\ &= \frac{(2n+1)!}{n!(n+1)!}\tag{1} \end{align}$$ And $$\begin{align} \frac 12{2n+2\choose n+1}&=\frac 12\frac{(2n+2)!}{(n+1)!(n+1)!} \\ &= \frac 12\frac{(2n+2)(2n+1)!}{(n+1)n!(n+1)!} \\ &=\frac{2n+2}{2(n+1)}\cdot\frac{(2n+1)!}{n!(n+1)!} \\ &= \frac{(2n+1)!}{n!(n+1)!}\tag{2} \end{align}$$ Finally, combine $(1)$ and $(2)$. You can also use Pascal's identity to immediately obtain the result in $(1)$.

bjorn93
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Say you have 2 red balls and other white. On how many ways you can choose $n+1$ of them? One answer is ${2n+2\choose n+1}$.

And second answer:

If both red are choosen then we must choose $n-1$ balls beetwen $2n$ balls, so that is $${2n\choose n-1}$$

If one red is choosen then we must choose $n$ balls beetwen $2n$ balls and one red, that is two ways, so that is $$2{2n\choose n}$$

If no red is choosen then we must choose $n+1$ balls beetwen $2n$ balls, so that is $${2n\choose n+1} = {2n\choose n-1}$$

Sum that and you are done.

nonuser
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